Mixing
Algebra with exponential functions
$begingroup$
If $f(x) = 4^x$ then show the value of $$f(x+1) - f(x)$$ in terms of $f(x)$.
I know the answer is $3f(x)$ because
$f(x+1)$ means that it is $4^x$ multiplied by 4 once more, which minus one is 3.
The question: How do I show this process algebraically? (Hints only please) I have tried using ln() functions to remove the powers to no avail.
$$ln(f(x)) = xln(4)$$
$$ln(f(x+1)) = (x+1)ln(4) = xln(4) + ln(4)$$
$$ln(f(x+1)) = ln(f(x)) + ln(4)$$
from here I don't know how to remove the natural logs to replace $f(x+1)$. What is a different approach I should use?
algebra-precalculus
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
$endgroup$
add a comment |
$begingroup$
If $f(x) = 4^x$ then show the value of $$f(x+1) - f(x)$$ in terms of $f(x)$.
I know the answer is $3f(x)$ because
$f(x+1)$ means that it is $4^x$ multiplied by 4 once more, which minus one is 3.
The question: How do I show this process algebraically? (Hints only please) I have tried using ln() functions to remove the powers to no avail.
$$ln(f(x)) = xln(4)$$
$$ln(f(x+1)) = (x+1)ln(4) = xln(4) + ln(4)$$
$$ln(f(x+1)) = ln(f(x)) + ln(4)$$
from here I don't know how to remove the natural logs to replace $f(x+1)$. What is a different approach I should use?
algebra-precalculus
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
$endgroup$
5
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35
add a comment |
$begingroup$
If $f(x) = 4^x$ then show the value of $$f(x+1) - f(x)$$ in terms of $f(x)$.
I know the answer is $3f(x)$ because
$f(x+1)$ means that it is $4^x$ multiplied by 4 once more, which minus one is 3.
The question: How do I show this process algebraically? (Hints only please) I have tried using ln() functions to remove the powers to no avail.
$$ln(f(x)) = xln(4)$$
$$ln(f(x+1)) = (x+1)ln(4) = xln(4) + ln(4)$$
$$ln(f(x+1)) = ln(f(x)) + ln(4)$$
from here I don't know how to remove the natural logs to replace $f(x+1)$. What is a different approach I should use?
algebra-precalculus
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
$endgroup$
If $f(x) = 4^x$ then show the value of $$f(x+1) - f(x)$$ in terms of $f(x)$.
I know the answer is $3f(x)$ because
$f(x+1)$ means that it is $4^x$ multiplied by 4 once more, which minus one is 3.
The question: How do I show this process algebraically? (Hints only please) I have tried using ln() functions to remove the powers to no avail.
$$ln(f(x)) = xln(4)$$
$$ln(f(x+1)) = (x+1)ln(4) = xln(4) + ln(4)$$
$$ln(f(x+1)) = ln(f(x)) + ln(4)$$
from here I don't know how to remove the natural logs to replace $f(x+1)$. What is a different approach I should use?
algebra-precalculus
algebra-precalculus
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
edited Jul 7 '18 at 8:13
Maria Mazur
49.9k1361125
49.9k1361125
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
asked Jul 7 '18 at 7:14
Po Chen LiuPo Chen Liu
1199
1199
5
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35
add a comment |
5
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35
5
5
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: $$f(x+1) = 4^{x+1} = 4^xcdot 4=...$$
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Here is a link to the solution, but maybe try to solve beforehand
Good luck :)
$$begin{align}
f(x+1)-f(x) &= 4^{x+1}-4^x \[1ex]
& = 4^xcdot 4-4^x \[1ex]
& = f(x)cdot 4-f(x) \[1ex]
& = 4f(x)-f(x) \[1ex]
& = 3f(x)
end{align}$$
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
$endgroup$
add a comment |
$begingroup$
Let $y = f(x)$. You want $f(x+1) - f(x)$ to be in terms of $y$, so you need to replace $x$ with an expression of $y$.
This expression is the inverse function of $f(x)$: $$x=f^{-1}(y)=log_4y$$
Therefore, $$f(x+1) - f(x)=f(f^{-1}(y)+1)-f(f^{-1}(y))$$
Simplify this expression to get $3y$.
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $$f(x+1) = 4^{x+1} = 4^xcdot 4=...$$
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Hint: $$f(x+1) = 4^{x+1} = 4^xcdot 4=...$$
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Hint: $$f(x+1) = 4^{x+1} = 4^xcdot 4=...$$
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
Hint: $$f(x+1) = 4^{x+1} = 4^xcdot 4=...$$
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
answered Jul 7 '18 at 7:15
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
add a comment |
add a comment |
$begingroup$
Here is a link to the solution, but maybe try to solve beforehand
Good luck :)
$$begin{align}
f(x+1)-f(x) &= 4^{x+1}-4^x \[1ex]
& = 4^xcdot 4-4^x \[1ex]
& = f(x)cdot 4-f(x) \[1ex]
& = 4f(x)-f(x) \[1ex]
& = 3f(x)
end{align}$$
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
$endgroup$
add a comment |
$begingroup$
Here is a link to the solution, but maybe try to solve beforehand
Good luck :)
$$begin{align}
f(x+1)-f(x) &= 4^{x+1}-4^x \[1ex]
& = 4^xcdot 4-4^x \[1ex]
& = f(x)cdot 4-f(x) \[1ex]
& = 4f(x)-f(x) \[1ex]
& = 3f(x)
end{align}$$
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
$endgroup$
add a comment |
$begingroup$
Here is a link to the solution, but maybe try to solve beforehand
Good luck :)
$$begin{align}
f(x+1)-f(x) &= 4^{x+1}-4^x \[1ex]
& = 4^xcdot 4-4^x \[1ex]
& = f(x)cdot 4-f(x) \[1ex]
& = 4f(x)-f(x) \[1ex]
& = 3f(x)
end{align}$$
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
$endgroup$
Here is a link to the solution, but maybe try to solve beforehand
Good luck :)
$$begin{align}
f(x+1)-f(x) &= 4^{x+1}-4^x \[1ex]
& = 4^xcdot 4-4^x \[1ex]
& = f(x)cdot 4-f(x) \[1ex]
& = 4f(x)-f(x) \[1ex]
& = 3f(x)
end{align}$$
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
edited Jul 7 '18 at 17:50
Rory Daulton
29.6k63355
29.6k63355
answered Jul 7 '18 at 10:31
AngAng
1718
answered Jul 7 '18 at 10:31
AngAng
1718
answered Jul 7 '18 at 10:31
AngAng
1718
1718
add a comment |
add a comment |
$begingroup$
Let $y = f(x)$. You want $f(x+1) - f(x)$ to be in terms of $y$, so you need to replace $x$ with an expression of $y$.
This expression is the inverse function of $f(x)$: $$x=f^{-1}(y)=log_4y$$
Therefore, $$f(x+1) - f(x)=f(f^{-1}(y)+1)-f(f^{-1}(y))$$
Simplify this expression to get $3y$.
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
$endgroup$
add a comment |
$begingroup$
Let $y = f(x)$. You want $f(x+1) - f(x)$ to be in terms of $y$, so you need to replace $x$ with an expression of $y$.
This expression is the inverse function of $f(x)$: $$x=f^{-1}(y)=log_4y$$
Therefore, $$f(x+1) - f(x)=f(f^{-1}(y)+1)-f(f^{-1}(y))$$
Simplify this expression to get $3y$.
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
$endgroup$
add a comment |
$begingroup$
Let $y = f(x)$. You want $f(x+1) - f(x)$ to be in terms of $y$, so you need to replace $x$ with an expression of $y$.
This expression is the inverse function of $f(x)$: $$x=f^{-1}(y)=log_4y$$
Therefore, $$f(x+1) - f(x)=f(f^{-1}(y)+1)-f(f^{-1}(y))$$
Simplify this expression to get $3y$.
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
$endgroup$
Let $y = f(x)$. You want $f(x+1) - f(x)$ to be in terms of $y$, so you need to replace $x$ with an expression of $y$.
This expression is the inverse function of $f(x)$: $$x=f^{-1}(y)=log_4y$$
Therefore, $$f(x+1) - f(x)=f(f^{-1}(y)+1)-f(f^{-1}(y))$$
Simplify this expression to get $3y$.
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
answered Jul 7 '18 at 17:59
jingyu9575jingyu9575
77252
77252
add a comment |
add a comment |
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5
$begingroup$
You are overthinking this. Your "I know the answer" is the answer. The words that follow say, correctly, that $4^{x+1} - 4^x = 4 times 4^x - 4^x = 3 times 4^x = 3f(x)$.
$endgroup$
– Ethan Bolker
Jul 7 '18 at 13:12
$begingroup$
Oh dear, thank you for your time anyway!....
$endgroup$
– Po Chen Liu
Jul 8 '18 at 23:35